Problem: You are given three integers a, b, k.

Find two binary integers x and y (x ≥ y) such that 

  1. both x and y consist of a zeroes and b ones; 
  2. x - y (also written in binary form) has exactly k ones. 

You are not allowed to use leading zeros for x and y. 

Input

The only line contains three integers a, b, and k (0 ≤ a; 1 ≤ b; 0 ≤ k ≤ a + b ≤ 2 ⋅ 10^5) — the number of zeroes, ones, and the number of ones in the result.

Output

If it's possible to find two suitable integers, print "Yes" followed by x and y in base-2.

Otherwise print "No".

If there are multiple possible answers, print any of them.

Examples

Input


4 2 3


Output


Yes
101000
100001


Input


3 2 1


Output


Yes
10100
10010


Input


3 2 5


Output


No

Note

In the first example, x = 101000_2 = 2^5 + 2^3 = 40_{10}, y = 100001_2 = 2^5 + 2^0 = 33_{10}, 40_{10} - 33_{10} = 7_{10} = 2^2 + 2^1 + 2^0 = 111_{2}. Hence x-y has 3 ones in base-2.

In the second example, x = 10100_2 = 2^4 + 2^2 = 20_{10}, y = 10010_2 = 2^4 + 2^1 = 18, x - y = 20 - 18 = 2_{10} = 10_{2}. This is precisely one 1.

In the third example, one may show, that it's impossible to find an answer.

Solution (in PYTHON3):

Solution:
import time,math as mt,bisect as bs,sys
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
from collections import defaultdict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
    return int(stdin.readline())
def IP(): # to take tuple as input
    return map(int,stdin.readline().split())
def L(): # to take list as input
    return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
    return stdout.write(str(x)+"\n")
def PI(x,y): # to print tuple separatedly
    return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
    return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
    if a==0:
        return b
    elif b==0:
        return a
    if a>b:
        return gcd(a%b,b)
    else:
        return gcd(a,b%a)
def bfs(adj,v): # a schema of bfs
    visited=[False]*(v+1)
    q=deque()
    while q:
        pass
def setBit(n):
    count=0
    while n!=0:
        n=n&(n-1)
        count+=1
    return count
mx=10**7
spf=[mx]*(mx+1)
def readTree(n,e): # to read tree
    adj=[set() for i in range(n+1)]
    for i in range(e):
        u1,u2=IP()
        adj[u1].add(u2)
    return adj
def sieve():
    li=[True]*(10**3+5)
    li[0],li[1]=False,False
    for i in range(2,len(li),1):
        if li[i]==True:
            for j in range(i*i,len(li),i):
                li[j]=False
    prime,cur=[0]*200,0
    for i in range(10**3+5):
        if li[i]==True:
            prime[cur]=i
            cur+=1
    return prime
spf=[1e9 for i in range(10**6+1)]
def SPF():
    mx=(10**6+1)
    spf[1]=1
    for i in range(2,mx):
        if spf[i]==1e9:
            spf[i]=i
            for j in range(i*i,mx,i):
                if i<spf[j]:
                    spf[j]=i
    return
def prime(n):
    prm=deque()
    d={}
    while n!=1:
        prm.append(spf[n])
        n=n//spf[n]
    for ele in prm:
        d[ele]=d.get(ele,0)+1
    return d
#####################################################################################
mod=998244353
inf = 10000000000000000
def solve():
    a,b,k=L()
    c=k+1
    if k==0:
        s=['0' for i in range(a+b)]
        for i in range(b):
            s[i]='1'
        ss="".join(s)
        print("Yes")
        print(ss)
        print(ss)
        return

    if a==0:
        print("No")
        return

    if (a+b-k-1)>0 and b>1:
        print("Yes")
        s=['0' for i in range(a+b)]
        t=['0' for i in range(a+b)]
        s[a+b-k-1]='1'
        t[a+b-1]='1'
        cnt,i=b-1,0
        while cnt:
            if s[i]=='0':
                s[i]='1'
                cnt-=1
            i+=1

        cnt,i=b-1,0
        while cnt:
            if t[i]=='0' and i!=(a+b-k-1):
                t[i]='1'
                cnt-=1
            i+=1
        ss="".join(s)
        tt="".join(t)
        print(ss)
        print(tt)
    else:
        print("No")
    return

t=1
for i in range(t):
    solve()

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